If $A=\begin{bmatrix}1&-2&3\\-4&

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Matrices

Question:

If $A=\begin{bmatrix}1&-2&3\\-4&2&5\end{bmatrix}$ and $B=\begin{bmatrix}2&3\\4&5\\2&1\end{bmatrix}$, then find the product AB and BA.

Options:

$AB=\begin{bmatrix}0&-4\\10&3\end{bmatrix},BA=\begin{bmatrix}-10&2&21\\-16&2&37\\-2&-2&11\end{bmatrix}$

$AB=\begin{bmatrix}0&-4\\16&3\end{bmatrix},BA=\begin{bmatrix}-10&2&11\\-16&2&37\\-2&-2&11\end{bmatrix}$

$AB=\begin{bmatrix}12&-4\\16&3\end{bmatrix},BA=\begin{bmatrix}-10&2&11\\-16&12&37\\-2&-2&11\end{bmatrix}$

None of these

Correct Answer:

$AB=\begin{bmatrix}0&-4\\10&3\end{bmatrix},BA=\begin{bmatrix}-10&2&21\\-16&2&37\\-2&-2&11\end{bmatrix}$

Explanation:

$AB=\begin{bmatrix}1&-2&3\\-4&2&5\end{bmatrix}\begin{bmatrix}2&3\\4&5\\2&1\end{bmatrix}$

$=\begin{bmatrix}2-8+6&-3-10+3\\-8+8+10&-12+10+5\end{bmatrix}$

$=\begin{bmatrix}0&-4\\10&3\end{bmatrix}$

$BA= \begin{bmatrix}2&3\\4&5\\2&1\end{bmatrix}\begin{bmatrix}1&-2&3\\-4&2&5\end{bmatrix}$

$=\begin{bmatrix}2-12&-4+6&6+15\\4-20&-8+10&12+25\\2-4&-4+2&6+5\end{bmatrix}$

$=\begin{bmatrix}-10&2&21\\-16&2&37\\-2&-2&11\end{bmatrix}$